Problem: Let $g$ be a polynomial function and let $g'$, its derivative, be defined as $g'(x)=x^5(x+1)(x-1)$. At how many points does the graph of $g$ have a relative minimum ? Choose 1 answer: Choose 1 answer: (Choice A) A None (Choice B) B One (Choice C) C Two (Choice D) D Three
Explanation: We can find the relative extrema (i.e. minima and maxima) of $g$ by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $g'(x)=x^5(x+1)(x-1)$. $g'(x)=0$ for $x=-1,0,1$. Since $g'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-1$, $x=0$, and $x=1$. $g$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into four intervals: $\llap{-}2.5$ $\llap{-}1.5$ $\llap{-}0.5$ $0.5$ $1.5$ $2.5$ $(-\infty,\ \ \llap{-}1)$ $(\ \ \llap{-}1,0)$ $(0,1)$ $(1,\infty)$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $(-\infty,-1)$ $x=-2$ $g'(-2)=-96<0$ $g$ is decreasing $\searrow$ $(-1,0)$ $x=-\dfrac12$ $g'\left(-\dfrac12\right)=\dfrac{3}{128}>0$ $g$ is increasing $\nearrow$ $(0,1)$ $x=\dfrac12$ $g'\left(\dfrac12\right)=-\dfrac{3}{128}<0$ $g$ is decreasing $\searrow$ $(1,\infty)$ $x=2$ $g'(2)=96>0$ $g$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $-1$ $\searrow$ $\nearrow$ Minimum $0$ $\nearrow$ $\searrow$ Maximum $1$ $\searrow$ $\nearrow$ Minimum Now we can see that $g$ has two relative minima.